Show me the steps to solve $($C$)$ Implement the algorithm In $[75]$: class TopKHeap$()$:

# The constructor of the class to initialize an empty data structure

def $($self$,$ k$)$:

$1$ self.k $=$ k

self. $A=[]$

self.H $=$ MinHeap$()$

def size$($self$)$:

return len$($self$.$A$)+($self$.$H$.$size$())$

def get$\_$jth$\_$element$($self$,$ j$)$:

assert $\frac{?}{\text{b}\text{a}\text{r}}(0)j$ self.k$-1$

assert $j$ self.size$()$

return self.A$[$j$]$

def satisfies$\_$assertions$($self$)$:

# is self.A sorted

for i in range$($len$($self$.A)$:

assert self.A$[$i self.A$[$i$+1],$ f'Array A fails to be sorted at position $\{$i$\},\{$self$.$A$[$i$],$ self.A$[$i$+1]\}\text{'}$

# is self.H a heap $($check min$-$heap property$)$

self.H$.$satisfies$\_$assertions$()$

# is every element of self.A less than or equal to each element of self.H

for $i$ in range$($len$($self$.$A$))$:

assert self.A$[$i self.H$.$min$\_$element$(),$ f'Array element $A[\{i\}]=\{self.A[i]\}$ is larger than min heap e

# Function : insert$\_$into$\_$A In $[75]$:

lass to initialize an empty data structure

self.H$.$size$())$

j$)$:

$-1$

elf$)$:

f$.$A$)-1)$:

$=$ self.A$[$i$+1],$ f'Array A fails to be sorted at position $\{$i$\},\{$self$.$A$[$i$],$ self.A$[$i$+1]\}\text{'}$

eck min$-$heap property$)$

tions$()$

self.A less than or equal to each element of self.H

f$.$A$))$:

$=$ self.H$.$min$\_$element$(),$ f'Array element A$[\{$i$\}]=\{$self$.$A$[$i$]\}$ is larger than min heap element $\{$self$.$H$.$min$\_$element$()\}\text{'}$

n that inserts an element into $.$

end of the array A

u just added at the very end of

per place so that the array A is sorted.

st element" jHat $($None if no element was displaced$)$

t$)$: def insert$\_$into$\_$A$($self$,$ elt$)$:

k $=$ self.k

print $("$k $=",$ k$)$

ass$$rt$($self$.$size$()$ k$)$

self.A$.$append$($elt$)$

j $=$ len$($self$.$A$)-1$

while $($j $>=1$ and self.A$[$j$]$ self.A$[$j$-1])$:$($self$.$A$[$j$],$ self.A$[$j$-1])=($self$.$A$[$j$-1],$ self.A$[$j$])$

j $=$ j $-1$

returnCode to handle when self.size $$ self.k is already provided

def insert$($self$,$ elt$)$:

size $=$ self.size$()$if size self.k:

self.insert$\_$into$\_$A$($elt$)$

returnyour code hereIn particular delete the j$^\{$th$\}$ element where j $=0$ means the least element.def delete$\_$top$\_$k$($self$,$ j$)$:

k $=$ self.k

assert self.size$()>$ k # we need not handle the case when size is less than or equal to $\}$

assert j $>=0$

assert j $$ self.k

Complete the implementation for maxheap data structure. First complete the implementation of MaxHeap. You can cut and paste relevant parts from

previous problems although we do not really recommend doing that. A better solution would have been to write a single implementation that could have

served as $mi\frac{n}{m}ax$ heap based on a flag.